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De Moivre's Theorem & Complex Numbers In Polar Form



in this article we're going to talk about complex numbers in polar form and we're also going to go over de moivre's theorem or de morpheus theorem the first thing you need to do is you need to be able to graph a complex number so this is going to be z in a plus b i format a is the real number bi is the imaginary number the x-axis is equivalent to the real axis and the imaginary axis is equivalent to the y-axis so let's say if we want to plot the complex number z equals 2 plus 3i we need to travel two units on the real axis and up three units on the imaginary axis so here is the first point let's call the first point a now to your turn plot this point negative three minus five I feel free to pause the video and plot it let's call that point b the real number is negative three so we gotta travel three to the left the imaginary number is negative 5i so we have to go down five units on the imaginary axis and so it should be somewhere in that vicinity so that's point b now how can we plot the complex number negative four what do you think we need to put that negative four is equivalent to negative four plus zero i so we gotta plot it at negative four zero that's four units to left on the real axis so it's going to be over here let's call that point c so that's z which is equal to negative 4. now what if z is equal to positive 2i where can we plot that so this is equivalent to zero plus two I which is the point zero comma two so basically that's on the y axis or the imaginary axis at two and let's call it pointing so that's how you can plot complex numbers in a complex plane now the next thing you need to be able to do is to find the absolute value of a complex number the absolute value of z let me write that better is equal to the absolute value of a plus bi which is equal to the square root of a squared plus b squared so if you think about it let's say if you plot a and bi let's say if you travel a units on the real axis and b units parallel to the imaginary axis z is basically the hypotenuse of that right triangle that's the magnitude of z or the absolute value of z so that equation is associated with the pythagorean theorem for a right triangle now let's say if z is 4 plus 3i let's find the absolute value of z so a is equal to four and b is equal to three so it's going to be the square root of four squared plus three squared four squared is 16 three squared is not 16 plus 9 is 25 and the square root of 25 is 5. so that's the absolute value of 4 plus 3i so now it's your turn try these problems let's say z is 4 minus 6i and also negative 1 plus 7i go ahead and calculate the absolute value of these two complex numbers so starting with the first one it's going to be the square root of four squared plus negative six squared four squared is 16 negative six times negative 6 is positive 36 16 plus 36 is 52 now this is not the final answer I mean it's correct but we could simplify the square root of 52. so think of a perfect square that goes into 52 perfect squares include numbers like 1 4 9 16 36 and so forth 4 is the highest perfect square that goes into 52 4 times 13 is 52 and the square root of 4 is 2. so the absolute value of 4 minus 6i is 2 root 13. now let's try the next one the absolute value of z is going to be the square root of negative 1 squared plus 7 squared negative 1 squared is 1 7 squared is 49 so this adds up to 50. now perfect square that goes into 50 is 25. 25 times 2 is 50 and the square root of 25 is 5. so root 50 simplifies to 5 root 2. and so that's how you can find the absolute value of z or any complex number now you need to know the difference between a complex number in rectangular form and a complex number in polar form what do you think the difference is between these two in rectangular form you have the letters a and b in polar form you have the letters r and theta so in rectangular form z is equal to a plus b I in polar form z is equal to r cosine theta plus I sine theta now rectangular form we mentioned that the x component of the right triangle is a the y component is b and the hypotenuse is the absolute value of z and polar form is very similar the legs of the triangle is still a and b the hypotenuse is r so r which is equivalent to the absolute value of z can be found using this equation some run out of space so r is going to be the square root of a squared plus b squared and to find the angle it's the inverse tangent of b divided by a but you need to be careful in the way in which you use that formula so let's work on some examples let's convert the following numbers from polar I mean from rectangular form to polar form so let's say z is equal to 3 plus 3i in rectangular form let's convert it to polar form by using this equation feel free to try we need to write it this way so first we got to find r so it's going to be the square root of a squared plus a b squared which they're both three three squared is nine and nine plus nine is 18 eighteen is basically nine times two and the squa re root of nine is three so r is three root two now we need to find the angle theta theta is going to be the arc tangent of b divided by a so that's 3 divided by 3 which is basically 1. the arc tangent of 1 is 45 degrees now this equation gives you the reference angle in the triangle however because z is in quadrant one a is three and b is string so r and z are both in quadrant one this angle is 45 we don't have to adjust it in any way i'll give you some other examples in which we need to adjust the angle so to write our final answer z is equal to 3 root 2 times cosine 45 plus I sine 45 if you want it in degrees now if you need to express your answer in radians you need to know that 45 is pi over 4. to convert it to radians multiply by pi over 180 180 divided by 45 is 4 so this becomes pi over 4. so you could say z is 3 root 2 cosine pi over 4 plus i sine pi over 4. but for the most part i'm going to represent the answer in degrees but if you want to you can convert it to radians using this process now let's work on some more examples let's say z is equal to 2 minus 2 root 3 I go ahead and try this one find r and then find theta now you want to find the angle relative from the positive x-axis not just a reference angle so let's start with r r is going to be the square root of a squared where a is 2 and b squared b is going to be 2 root 3. now what is 2 root 3 squared so that's basically two root three times two root three two times two is four and root three times root three is the square root of nine which is three and four times three is 12 two squared is four and two v three squared is 12 four plus 12 is 16 so we got the square root of 16 so r is four so now that we have the value of r let's find a reference angle to begin with to find a reference angle which is the angle inside the triangle when you use this equation make sure that a and b are positive so in this example b even though it's a negative 2 root 3 i'm going to use positive 2 3 in the equation and a is 2. so what we're looking for is the arc tangent of root 3. if you type that in your calculator make sure it's in degree mode you should get 60 degrees that's the reference angle now z is not in quadrant one so the final answer is not 60 degrees notice that a is positive which is directed towards the right in the positive x-axis but b is negative so therefore r is in quadrant four so this is positive two and this is negative two root three now the reference angle is the angle inside the triangle that's sixty but we want the angle measured from the positive x-axis so that's 360 minus 60 which is 300 that's the angle that we want to report now here's some notes that you may want to keep in mind let's say if r is in quadrant one to find the angle is the same as the reference angle you don't need to change it in quadrant two your angle has to be between 90 and 180 so the angle in quadrant two it's 180 minus the reference angle if r is in quadrant three the angle is going to be 180 plus the reference angle and anytime r is in quadrant 4 it's going to be 360 minus the reference angle which is what we did 360 minus 60 and that gave us this angle which is 300 so now we can write our final answer in polar form so using equation z is equal to r cosine theta plus I sine theta z is going to be 4 cosine 300 plus i sine 300 and if we want to write it in radians let's convert 300 so let's multiply by pi divided by 180 so we could cancel zero so we have 30 pi over 18 thirty is basically six times five 18 is six times three so we can get rid of a six so therefore three hundred degrees is five pi over three so this can be written as four cosine five pi over three plus i sine five pi over three here's another example that you could try let's say z is equal to negative three plus five i go ahead and convert it into its polar form take a minute pause the video and try this problem so let's start by finding r r is the square root of a squared plus b squared so negative three squared is nine five squared is twenty-five nine plus twenty-five is 34 and we can't simplify root 34 because there's no perfect square that goes into 34. all you could do is break down 34 into 2 and 17 and both of these numbers are prime numbers so now let's find the reference angle so it's going to be arc tangent b over a and we're just going to use the positive values of b and a so our tan 5 divided by 3 which is 59 degrees it's 59.03 but i'm going to round it to 59 degrees now let's find out what quadrant r is located in so a is negative 3 that means we're going to travel 3 units to the left and b is positive 5. so we've got to go up 5 units which means r is in quadrant two anytime you have r in quadrant two the angle is going to be 180 minus the reference angle which is 59 so 180 minus 59 is 121. this is the angle that we want relative to the positive x-axis that's 121. the reference angle is inside the triangle that's 59. these two have to add up to 180 which is a straight line so now that we have the angle we can write our final answer in polar form so z is equal to r which is the square root of 34 times cosine of 121 plus i sine 121 and that's in degrees so far we've considered examples of converting complex numbers in rectangular form to polar form if we're given a and b now what if we're given a or b but not both what should we do for example let's say if z is equal to three how do you write that in polar form well if z is equal to just one number if you have a or b only r is going to be equal to z it's just going to be a or b by itself so in this case r is string to prove it when z is 3 is the same as being 3 plus 0 i so to find r it's going to be the square root of a squared plus the b squared 3 squared is 9 plus 0 squared that's still going to be 9 and the square root of 9 is 3. so r is 3. so if you're given a or b alone and the other one is 0 then z is equal to r under such circumstances now we need to find the angle and it won't be wise to use the inverse tangent formula if you type in inverse tangent y over x you get arc tan zero which may give you the right answer in this case it will it's zero but the easiest way to do it is to plot z three plus zero I is on the x-axis it's over here so if you draw r notice that r is three units long that's why r equals three and it's at an angle of zero degrees so if you have a or b and not both the angles is either going to be 0 90 180 or 270. because z is equal to positive 3 the angle is 0 degrees so therefore we could say z is equal to r which is three times cosine of zero degrees plus i sine of zero degrees so that's c in polar form now what if z was equal to negative four r is always a positive value that means r is equal to four so this is negative four plus zero i r would be the square root of negative 4 squared plus 0 squared which is the square root of 16 which is 4. now if we try using the inverse tangent formula if you type this in you'll get zero degrees but the angle is not zero because this time if you plot it negative 4 0 I is on the negative x axis it's 4 units to the left so r is 4 as you can see but the angle is not zero rather the angle is 180 which is on the negative x axis so that's why this formula doesn't always work if you have a or b but not both it's better to graph it just to make sure so the answer for this problem is 4 cosine of 180 degrees plus I sine 180 try these two problems let's say that z is negative 2i and also that z is positive 5i go ahead and try those two so first we got to find r you can see that r is equal to two keep in mind z being negative two I is zero minus two I so r would be the square root of zero squared plus negative two squared which is the square root of four which is two so r is simply two now let's not use the inverse tangent formula instead let's just plot negative 2i negative 2i is on the negative y-axis so you can see r is 2 units long and the angle at this point is 270 degrees so therefore z is equal to r which is two times cosine of 270 plus i sine 270. now i'll try the other one z is equal to 5i so clearly we see that r is 5. if we plot it we just got to travel 5 units in the y direction and we can see that the angle in the positive y axis is 90. so therefore z is going to be r which is 5 cosine of ninety plus I sine ninety now we need to know how to go back from polar form into rectangular form and this method is pretty straightforward so let's see if we have 4 cosine 90 plus I sine 90. convert it back to rectangular form so what is cosine 90 cosine 90 is 0 and sine 90 is 1. 4 times 0 is 0 and 4 times 1i is 4i so therefore z is equal to 4i and that's all you need to do to put it back into rectangular form so let's work on some more problems try this one let's say that z is equal to 10. times cosine thirty plus i sine thirty cosine thirty is root three divided by two and sine thirty is one half now let's distribute 10. 10 times root 3 over 2. 10 divided by 2 is 5. so this is going to be 5 square root 3 and half of 10 is 5 so plus 5i and that's the answer try this one let's say that z is equal to 20 cosine 4 pi over 3 plus i sine 4 pi over 3. let's convert the angle from radians to degrees so let's multiply by 180 over pi so pi cancels 18 divided by 3 is 6 so 180 divided by 3 is 60. 4 times 60 is 240 cosine 240 that's the same as cosine 60 60 is the reference angle of 240. if you do 240 minus 180 that's 60. cosine 60 is one half so cosine 240 is also one half but 240 is in quadrant three so cosine 240 is a negative one half sine 240 which is similar to sine 60 sine 60 is root 3 over 2 and sine 240 that's going to be negative root 3 over 2 because sine is negative in quadrant 3. half of 20 is 10 so this is going to be negative 10 minus 10 root 3i now for those of you out there who may need to evaluate cosine 240 or sine 240 without a calculator here's what you can do so first draw the angle 240. 240 is in quadrant three so this is 240 degrees and turn it into a triangle so this is 180 a nd if this is 240 the difference must be 60. so you need to realize that 60 is the reference angle to find the reference angle and if your angle is in quadrant three it's going to be the angle in quadrant three minus 180 to find it based on the angle in quadrant two it's 180 minus the angle in quadrant two and if you have an angle in quadrant four it's going to be 360 minus the angle in quadrant four now that we have the reference angle we can solve it using the 30 60 90 reference triangle across the 30 is always going to be one across the 60 root 3 across the 92 so now we can evaluate sine of 60. sine based on sohcahtoa is equal to the opposite side divided by the hypotenuse opposite to 60 is root 3. the hypotenuse is across the box so that's going to be 2. so sine 60 is root 3 over 2. now sine 240 will also be root 3 over 2. you just got to determine if it's positive or negative sine is associated with the y value and y is negative in quadrant three so this is going to be negative root three over two cosine 240 is similar to cosine 16 cosine is adjacent over hypotenuse adjacent to 60 is one the hypotenuse is two so now we can find cosine 240. in quadrant three x is negative so cosine 240 is negative therefore it's negative one half and so that's how you could find these values based on the 30-60-90 triangle our next topic of discussion is finding the product of two complex numbers so for example let's say that z1 is r1 cosine theta 1 plus i sine theta 1 and z2 is r2 cosine theta 2 plus i sine theta 2. what is z1 times z2 what is the product of these two complex numbers the product is simply going to be r1 times r2 and then it's going to be cosine of the sum of the two angles that is theta 1 plus theta 2 plus I sine theta 1 plus theta 2. so all we need to do to multiply z1 and z2 we just got to multiply r1 and r2 whatever two numbers are in front we got to multiply them and then we just got to add the angles theta 1 plus theta 2. so let's try some examples let's say that z1 is equal to 5 cosine 30 plus i sine 30 and that z2 let's say is equal to 7 cosine 45 plus i sine 45 what is z1 times z2 so all we got to do is multiply 5 times 7 which is 35 and then we need to add the angles 30 plus 45 is 75 so it's going to be cosine 75 degrees plus I sine 75 degrees and that's it here's the practice problem that you could try let's say that z1 is 8 cosine 60 plus I sine 60 and also that z2 is 5 cosine 100 plus I sine 100 go ahead and find the product of z1 and z2 so feel free to pause the video and work on it so first let's multiply eight times five so that's going to be 40. next let's add 60 plus 100 which is going to be 160. so the answer is 40 cosine 160 plus i sine 160. 

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